I am struggling to see how to solve this question.
My confusion stems from the Less probable, for n plain drones the equation I derived was $9\times10^{\frac{n+1}{2}-1}$ this is under the assumption that the plaindrone dose not start with a zero. So I set up the equation to see if I could get a feeling, but still not sure.
$$\frac{9 \times 10^{\frac{n+1}{2}-1}}{n}<6.7\times10^{-20} $$
So I thought of a dice and the say the probability of rolling a $1$ $p=\frac{1}{6}$ is less probable that rolling a $2$ $3$ or a $4$ for $p=\frac{3}{6}$. But this still didn't really give me any insight to this question. Just wondering if someone could expand what less probable means in this context.
Let's use this notation:
$$x=10^na_n+10^{n-1}a_{n-1}+...+10a_1+a_0=\overline{a_na_{n-1}...a_1a_0}$$
A number of $n$ digits is palyindromic if and only if $$a_{n-k}=a_k \ \ \forall k<n$$
This means that we have to choose $\lceil{\frac n2}\rceil$ ciphers(the others one are forced). All of this ciphers can vary between $0$ and $9$ except for $a_n\neq 0$ so:
$$ 9\times 10^{\lceil{\frac n2}\rceil-1}$$
That can be rewritten as:
$$9\times 10^{\frac n2 -1} \ \ \ \text{if} \ \ n \ \ \text{is even}$$ $$9\times 10^{\frac {n+1}{2} -1} \ \ \ \text{if} \ \ n \ \ \text{is odd}$$
Clearly all of the possible numbers of $n$ digits are $9\times 10^{n-1}$ so our probability will be(dividing good cases by possible cases):
$$10^{-\frac n2 } \ \ \ \text{if} \ \ n \ \ \text{is even}$$ $$ 10^{\frac {1-n}{2} } \ \ \ \text{if} \ \ n \ \ \text{is odd}$$
Formally we should do $2$ disequations. Indicating with $p$ the probabily of the neutrino. If $n$ is even:
$$10^{-\frac n2 }<p$$ $$-\frac n2 <\log_{10}(p)$$ $$n>-2\log_{10}(p)$$ $$n>\log_{10}(\frac{1}{p^2})$$ $$n>38.34$$ $$n\geq 40$$
If $n$ is odd :
$$10^{\frac 12-\frac n2 }<p$$ $$\frac 12-\frac n2 <\log_{10}(p)$$ $$n>1-2\log_{10}(p)$$ $$n>1+\log_{10}(\frac{1}{p^2})$$ $$n>39.34$$ $$n\geq 41$$
So $n=40$ is enough :).